The number of distinct values of `lamda`, for which the vectors `-lamda^(2)hati+hatj+hatk, hati-lamda^(2)hatj+hatk` and `hati+hatj-lamda^(2)hatk` are coplanar, is

Options

A. zero
B. one
C. two
D. three

Explanation

To determine the number of distinct values of \( \lambda \) for which the vectors \( -\lambda^2 \hat{i} + \hat{j} + \hat{k} \), \( \hat{i} - \lambda^2 \hat{j} + \hat{k} \), and \( \hat{i} + \hat{j} - \lambda^2 \hat{k} \) are coplanar, we will follow these steps:

### Step 1: Define the Vectors
Let:
- \( \mathbf{A} = -\lambda^2 \hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{B} = \hat{i} - \lambda^2 \hat{j} + \hat{k} \)
- \( \mathbf{C} = \hat{i} + \hat{j} - \lambda^2 \hat{k} \)

### Step 2: Set Up the Scalar Triple Product
The condition for coplanarity of the three vectors is that their scalar triple product is zero:
\[ \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 0 \]
This can be calculated using the determinant of a matrix formed by the components of the vectors.

### Step 3: Form the Determinant
We can express the scalar triple product as the determinant:
\[ \Delta = \begin{vmatrix} -\lambda^2 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2 \end{vmatrix} \]

### Step 4: Calculate the Determinant
We will compute the determinant \( \Delta \):
1. Expand the determinant using cofactor expansion or row operations.
2. Perform row operations to simplify the determinant.

After performing the necessary operations, we get:
\[ \Delta = (2 - \lambda^2)(1 + \lambda^2)^2 \]

### Step 5: Set the Determinant to Zero
For coplanarity, we set the determinant to zero:
\[ (2 - \lambda^2)(1 + \lambda^2)^2 = 0 \]

### Step 6: Solve for \( \lambda \)
This equation gives us two factors to consider:
1. \( 2 - \lambda^2 = 0 \) which leads to \( \lambda^2 = 2 \) or \( \lambda = \pm \sqrt{2} \).
2. \( (1 + \lambda^2)^2 = 0 \) which has no real solutions since \( 1 + \lambda^2 \) is always positive for real \( \lambda \).

### Step 7: Count Distinct Values
The only distinct values of \( \lambda \) that satisfy the coplanarity condition are:
- \( \lambda = \sqrt{2} \)
- \( \lambda = -\sqrt{2} \)

Thus, there are **2 distinct values** of \( \lambda \).

### Final Answer
The number of distinct values of \( \lambda \) for which the vectors are coplanar is **2**.

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